Monty Hall Problem

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Can someone explain why the new probablilities of winning are not 1/2? I've read every article on this problem and still don't understand it! It's a paradox!

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Can someone explain why the new probablilities of winning are not 1/2? I've read every article on this problem and still don't understand it! It's a paradox!

I can explain it (at least the way I was told) but I'm not sure I can convince anyone because I'm not convinced myself.

Let's change this from 3 doors to, say, 10.
You pick, say, door 5.
Monty says, "I'll tell you it's not behind door 7. Do you want to make a new choice?" You say no.

On and on it goes until there are two doors left, door 3 and door 5.

It is better to pick door 3 as your final choice.

The reason is simple enough: Your original choice of a door had a 1 in 10 chance of being correct. Your final choice has a 1 in 2 chance of being correct. The odds are with the other door being right.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Can someone explain why the new probablilities of winning are not 1/2? I've read every article on this problem and still don't understand it! It's a paradox!

I love this classic problem.

Think of this, if you use the strategy of chosing the other door then let us analyze what happens. If you pick the wrong door out of the three then you win. HOW?? Because if you pick the wrong door the other bad is revealed and hence the good door remains. But the probability of selecting a bad door is 2/3 thus, the probability of success is also 2/3. You can also think of it like this. If you pick the good door then you lose. WHY? Because the bad door is revealed and by changing you pick the bad door. Now the probability of the good door is 1/3. Thus, the probability of failure is 1/3, which means the probability of winning is 1-1/3=2/3.
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Another explanation. Since there are equal chances each door is of probability 1/3. Hence, when one door is open and revealed its probability is now zero, because it is definelty bad. But the door you chose is of 1/3 hence the remaining door is 2/3 (sum of probabilities is one i.e. 0+1/3+x=1).
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Think of this. There are 1,000,000,002 doors. You chose one (highly unlikey to win) and the other billion are shown to be false and only one remains. Do you really think by staying the probability is 1/2? No way!
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We can generalize this as, if there are doors then using this strategy the probability is

Think of this, if you use the strategy of chosing the other door then let us analyze what happens. If you pick the wrong door out of the three then you win. HOW?? Because if you pick the wrong door the other bad is revealed and hence the good door remains. But the probability of selecting a bad door is 2/3 thus, the probability of success is also 2/3. You can also think of it like this. If you pick the good door then you lose. WHY? Because the bad door is revealed and by changing you pick the bad door. Now the probability of the good door is 1/3. Thus, the probability of failure is 1/3, which means the probability of winning is 1-1/3=2/3.

But the contestant doesn't know if he picked the wrong door or not.

Originally Posted by ThePerfectHacker

Think of this. There are 1,000,000,002 doors. You chose one (highly unlikey to win) and the other billion are shown to be false and only one remains. Do you really think by staying the probability is 1/2? No way!

What!? How are your chances of winning not not 1/2? There are only 2 doors remaining. One door wins and the other door loses. This is really counterintuitive. Is there some kind of math proof that you can share to show that your chances are not 1/2?

think about it like this

Here is the way I learnt to view the problem:

Put yourself in the position of the host:

The contestant has just picked a door, and you must now eliminate another door.
Now, since 2 out of 3 doors are incorrect you would agree that MOST of the time- you (the host) will be left with ONE incorrect door which you HAVE to eliminate hence leaving only the remaining correct door.
It follows therefore that if the contestant switches, then MOST of the time he will be left with the remaining correct door.

(Of course, the contestant may by chance have picked the correct door to start with, but there is only 1/3 chance of this happening!)